Nice approach, but just because the function is continuous, doesn't mean that it's differentiable.

Lets say we have f(x) > 0 and f(y) < 0 with x,y on the interval.
let a = min (x,y) and b=max(x,y)
Clearly (a,b) is contained in the interval. Therefore f is continuous on [a,b].
It's also clear that 0 is between f(a) and f(a)

Then by applying the intermediate value theorem we get hat there must be some point x on the interval [a,b] such that f(x)=0, but this contradicts the hypothesis that [tex]f(x) \neq 0[/tex] on the interval.

How you would do this depends upon what prior theorems you have to work with. Have you had the "intermediate value theorem"? That says that if f(x) is continuous on [a,b] then, on [a,b], f takes on all values between f(a) and f(b).